A few families took a trip to an amusement park together. Tickets cost $$6.50$ each for adults and $$4.50$ each for kids, and the group paid $$35.50$ in total. There were $3$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Solution: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${6.5x+4.5y = 35.5}$ ${x = y-3}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-3}$ for $x$ in the first equation. ${6.5}{(y-3)}{+ 4.5y = 35.5}$ Simplify and solve for $y$ $ 6.5y-19.5 + 4.5y = 35.5 $ $ 11y-19.5 = 35.5 $ $ 11y = 55 $ $ y = \dfrac{55}{11} $ ${y = 5}$ Now that you know ${y = 5}$ , plug it back into ${x = y-3}$ to find $x$ ${x = }{(5)}{ - 3}$ ${x = 2}$ You can also plug ${y = 5}$ into ${6.5x+4.5y = 35.5}$ and get the same answer for $x$ ${6.5x + 4.5}{(5)}{= 35.5}$ ${x = 2}$ There were $2$ adults and $5$ kids.